Define the mapping $\,\phi(z) := z^2+c\,$ with fixed points
$\,m\,$ and $\,M\,$ which famously occurs in the definition of the
Mandelbrot set.
The problem of how to numerically interpolate the
iterated function
(also referred to as
iterated composition)
of $\,\phi(z)\,$ is just one special case in the area of
complex dynamics.
To begin, define a recursive sequence with
$$ a_0 := a,\qquad a_n = \phi^{\circ n}(a) := \phi(a_{n-1}). \tag1 $$
Define notation $\, q := 2m\,$ and verify that
$\, c = m\, M, \; 1 = m+M .\,$ Let
$$ T(x) := q\,x(1-x),\;\; K(x) := m - q\,x,\;\;
K^{\circ -1}(x) = \frac{m-x}q. \tag2 $$
Verify that
$$ x = K^{\circ -1}\circ K(x) := K^{\circ -1}(K(x)) $$
and
$$ T(x) = K^{\circ -1}\circ\phi\circ K(x) :=
K^{\circ -1}(\phi(K(x))). \tag3 $$
In other words, the quadratic mapping $\,\phi\,$ is conjugate
via the linear map $\,K\,$ and its
inverse function
$\,K^{\circ -1}\,$ to a
logistic map
$\,T\,$ with fixed points at $\,0\,$ and $\,\infty.\,$
From Schröder's equation
there is a formal power series expansion
$$ g(z) := z + c_2\,z^2 + c_3\,z^3 + c_4\,z^4 + \cdots\tag4 $$
such that the functional equation
$$ q\,g(z) = g\circ T(z) := g(T(z)) \tag5 $$
is satisfied where the coefficients $\,c_2,\,c_3,\,c_4,\,\dots\,$
are determined by the functional equation.
Substitute definition $(4)$ into equation $(5)$ and equate
coefficients to solve for $\,c_2,\,c_3,\,c_4.\,$
The result is
$$ c_2 = \frac{1}{q\!-\!1},
c_3 = \frac{2q}{(q\!-\!1)(q^2\!-\!1)},
c_4 = \frac{q+5q^3}{(q\!-\!1)(q^2\!-\!1)(q^3\!-\!1)}.
\tag6 $$
Note that for the exceptional case of $\,q=1,\,m=M=1/2\,$
these coefficients are all $\,\infty,\,$ $T(z) = z(1-z)\,$
and $\,\phi(z) = z^2+1/4.\,$ This corresponds to the map
$\,x\to x-x^2\,$ which has a different convergence behavior
as in several MSE questions such as
MSE 2861768.
Apply the functional equation $(5)$ iteratively to get
$$ q^n\,g(z) = g(T^{\circ n}(z)),\;\;
T^{\circ n}(z) = g^{\circ -1}(q^n g(z)). \tag7 $$
Use equation $(3)$ to get
$$ \phi^{\circ n}(x) =
K\circ T^{\circ n}\circ K^{\circ-1}(x) :=
K(T^{\circ n}(K^{\circ-1}(x))). \tag8 $$
Now the $\,n\,$ can be replaced by any real number. For
example, with $\,n = 1/2,\,$ define
$$ \theta(m \!+\!z) := m \!+\!u\,z \!+\!\frac{z^2}{u+u^2}
\!-\!\frac{2z^3}{u^2(1+u)^2(1+u^2)} + \cdots \tag9 $$
where $\, u := \sqrt{q}.\,$ Then $\,\phi(z) =
\theta^{\circ2}(z) := \theta\circ\theta(z) :=
\theta(\theta(z)).\,$ Verify that
$$ \theta^{\circ2}(m \!+\!z) = m + 2mz + z^2 =
(m \!+\!z)^2 + c. \tag{10} $$
However, a more direct approach is to assume an Ansatz of
$$ \phi^{\circ n}(m\!+\!z) = m + c_1\,x + c_2\,x^2 +
c_3\,x^3 + \dots \tag{11}$$
where $\,c_1,\,c_2,\,c_3,\,\dots\,$ depend on $\,m\,$
and $\,n\,$ only. This formal power series must satisfy
the equation
$$ \phi\circ\phi^{\circ n}(m\!+\!z) :=
\phi(\phi^{\circ n}(m\!+\!z)) =
\phi^{\circ n+1}(m\!+\!z). \tag{12}$$
Substitute definition $(11)$ into equation $(12)$ and
equate coefficients to solve for
$\,c_1,\,c_2,\,c_3.\,$ The result is that
$$ c_1 = q^n, \; c_2 = \frac{q^n(q^n-1)}
{q(q-1)}, \; c_3 = \frac{2q^n(q^n-1)(q^n-q)}
{q^2(q-1)(q^2-1)}. \tag{13} $$
As expected, when $\,n=1\,$ equation $(13)$ reduces to
$$ c_1 = 2m,\quad c_2 = 1,\quad c_3 = 0,\quad\dots\,
\tag{14} $$
agreeing with equation $(10)$. Also, with $\,n=1/2\,$
equation $(13)$ agrees with the half iterate equation $(9)$.
The question asks for a closed-form function for the
$\,\phi^{\circ n}(x)$ of equation $(8)$
Is that even possible? If so, how do I find that function?
If not, are there alternatives?
The answer to this is no, except in rare cases. An alternative
is given by using the power series of $\,g(z)\,$ as given in
equation $(4)$ or directly the one in equation $(11)$.
